Quiz

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The remainder factor theorem is actually two theorems that relate the roots of a polynomial with its linear factors. The theorem is often used to help factorize polynomials without the use of long division. Especially when combined with the rational root theorem, this gives us a powerful tool to factor polynomials.

Remainder Theorem:

For a polynomial \( f(x)\), the remainder of \( f(x)\) upon division by \( x-c\) is \( f(c)\). \(_\square\)

Factor Theorem:

Let \( f(x)\) be a polynomial such that \( f(c) =0\) for some constant \( c\). Then \( x-c\) is a factor of \( f(x)\). Conversely, if \( x-c\) is a factor of \( f(x)\), then \( f(c)=0\). \(_\square\)

Contents

Remainder Factor Theorem - Basic

Completely factorize \( f(x) = 6x^3 - 23x^2 - 6x+8\).

From the rational root theorem, we try numbers of the form \( \frac {a}{b}\), where \( a\) divides 8 and \( b\) divides 6. By the remainder factor theorem, we just need to calculate the values for which \( f \left( \frac {a}{b} \right)=0\). Through trial and error, we obtain

Remainder Factor Theorem - Intermediate

\( g(x)\) is a polynomial that leaves a remainder of 1 when divided by \( x-1\) and leaves a remainder of 4 when divided by \( x+2\). What is the remainder when \( g(x)\) is divided by \( (x-1)(x+2)\)?

We have \( g(x) = (x-1)(x+2)q(x) + r(x)\). Since \( (x-1)(x+2)\) has degree 2, the remainder \( r(x)\) has degree at most 1, hence \( r(x)=Ax+B\) for some constants \( A\) and \( B\). By the remainder factor theorem, we have \( g(1)=1\) and \( g(-2)=4\). Substituting \( x=1\) and \( -2\), we obtain

Factor:

(Backward direction)
If \( x-c\) is a factor of \( f(x)\), then (by definition) the remainder of \( f(x)\) upon division by \( x-c\) would be 0. By the remainder theorem, this is equal to \( f(c)\). Hence, \( f(c)=0\). \( _\square \)

Note: There are no restrictions on the constant \( c\). It could be a real number, a complex number, or even a matrix!